Problem: The lifespans of sloths in a particular zoo are normally distributed. The average sloth lives $19$ years; the standard deviation is $3.2$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a sloth living between $9.4$ and $22.2$ years.
Solution: $19$ $15.8$ $22.2$ $12.6$ $25.4$ $9.4$ $28.6$ $99.7\%$ $68\%$ $15.85\%$ $15.85\%$ We know the lifespans are normally distributed with an average lifespan of $19$ years. We know the standard deviation is $3.2$ years, so one standard deviation below the mean is $15.8$ years and one standard deviation above the mean is $22.2$ years. Two standard deviations below the mean is $12.6$ years and two standard deviations above the mean is $25.4$ years. Three standard deviations below the mean is $9.4$ years and three standard deviations above the mean is $28.6$ years. We are interested in the probability of a sloth living between $9.4$ and $22.2$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the sloths will have lifespans within 3 standard deviations of the average lifespan. It also tells us that $68\%$ of the sloths will have lifespans within 1 standard deviation of the mean. The probability of a particular sloth living between $9.4$ and $22.2$ years is $\color{orange}{15.85\%} + {68\%}$, or $83.85\%$.